Question

Tanx/2=tany/3=tanz/5 and x+y+z=π then the value of tan^2x+tan^2y+tan^2z is

Answer 1

It has given that,

tanx/2 = tany/3 = tanz/5 and x + y + z = π

To find : The value of tan²x + tan²y + tan²z

solution : here, x + y + z = π

⇒tan(x + y + z) = tanπ

⇒{tanx + tany + tanz - tanx tany tanz}/{1 - tanx tany - tany tanz - tanz tanx } = 0

⇒tanx + tany + tanz = tanx tany tanz......(1)

now tanx/2 = tany/3 = tanz/5 = k (let)

⇒tanx = 2k , tany = 3k , tanz = 5k

putting above terms in equation (1) we get,

2k + 3k + 5k = 2k . 3k . 5k

⇒10k = 30k³

⇒k² = 1/3

⇒k = 1/√3

now, tanx = 2/√3, tany = 3/√3 and tanz = 5/√3

so, tan²x + tan²y + tan²z

= (2/√3)² + (3/√3)² + (5/√3)²

= (2² + 3² + 5²)/3

= (4 + 9 + 25)/3

= 38/3

Therefore the value of tan²x + tan²y + tan²z = 38/3